INTEGER-VALUED POLYNOMIALS on KRULL RINGS Introduction If A

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 124, Number 12, December 1996, Pages 3595–3604 S 0002-9939(96)03442-9

INTEGER-VALUED POLYNOMIALS ON KRULL RINGS

SOPHIE FRISCH

(Communicated by Wolmer V. Vasconcelos)

Abstract. If R is a subring of a Krull ring S such that RQ is a valuation ring for every finite index Q = P R, P in Spec1(S), we construct polynomials that map R into the maximal possible∩ (for a monic polynomial of fixed degree) 1 power of PSP, for all P in Spec (S) simultaneously. This gives a direct sum decomposition of Int(R, S), the S-module of polynomials with coefficients in the quotient field of S that map R into S, and a criterion when Int(R, S)has a regular basis (one consisting of 1 polynomial of each non-negative degree).

Introduction If A is an infinite subset of a domain S, we write Int(A, S)fortheS-module of polynomials with coefficients in the quotient field of S that – when acting as a function by substitution of the variable – map A into S.ForInt(S, S), the ring of integer-valued polynomials on S, we write Int(S). Beyond the fact (known of x x(x 1)...(x n+1) old) that the binomial polynomials n = − n! − form a basis of the free Z-module Int(Z), the study of Int(S)originatedwithPólya [16] and Ostrowski [15], who let S be the ring of integers in a number field (their results have been generalized to Dedekind rings by Cahen [4]). Int(R, S)forR=Shas only begun to attract attention more recently [2], [3], [6], [8], [11], [13]. 6 We will treat Pólya’s and Ostrowski’s questions in the case where R = S and S is a Krull ring; in particular the question when Int(R, S)isafreeS-module6 that admits a regular basis, and the related one of determining the highest power of PSP, where P is a height 1 prime ideal of S, that a monic polynomial of fixed degree can

map R into. Following Pólya, we call a sequence of polynomials (gn)n 0 regular, ∈N if deg gn = n for all n. One basic connection between a module of polynomials and the modules of leading coefficients should be kept in mind: 0.1 Lemma. Let R be a unitary subring of a field K, M an R-submodule of K[x], and In = leading coefficients of n-th degree polynomials in M 0 . { }∪{ } (i) If (gn)n 0 is a regular sequence of monic polynomials in K[x] such that ∈N Ingn M for all n,thenM= n∞=0 Ingn (direct sum). (ii) A regular⊆ set of polynomials in M is an R-basis if and only if the leading P coefficient of the n-th degree polynomial generates In as an R-module. (iii) M has a regular R-basis if and only if each In is non-zero and cyclic.

Received by the editors September 2, 1994 and, in revised form, May 1, 1995. 1991 Mathematics Subject Classiﬁcation. Primary 13B25, 13F05; Secondary 13F20, 11C08.

c 1996 American Mathematical Society

3595

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Proof. (i) If (gn)n 0 is as stated, then n∞=0 Ingn M and the sum is direct, since ∈N ⊆ deg(gn)=nmakes the gn linearly independent over K. An induction on N =degf N P shows that f M implies f Ingn. Indeed, for N =0,f I0 =g0I0,and ∈ ∈ n=0 ∈ if N>0andaN is f’s leading coefficient, then aN IN ,soh=f aNgN M N 1 P ∈ − ∈ and h − Ingn by induction hypothesis. (ii) and (iii) are easy. ∈ n=0 1.P Polynomials mapping a set into a discrete valuation ring Throughout section one, v is a discrete valuation on a field K with value-group Γv = Z and v(0) = ,andRv its valuation ring with maximal ideal Mv.Ina kind of generic local regular∞ basis theorem, we will establish the connection (well- known in special cases) between Int(A, Rv) and the maximal power of Mv that a monic polynomial of degree n can map A into, for all A K for which this maximum exists for every n. A subset A of the quotient field⊆ of a domain R is called R-fractional if there exists a d R 0 such that dA R. ∈ \{ } ⊆ 1.0 Lemma. If R is an integrally closed domain with quotient field L, A L and f non-constant L[x] then f(A) is R-fractional if and only if A is. ⊆ ∈ Proof. Let f L[x], deg f = n>0. If f(A)isR-fractional there is a non-zero d R,withdf∈ (a) R for every a A.Letc R 0 , such that cf R[x], ∈ ∈ n ∈ ∈ \{ } ∈ and set g = cdf = cnx + ...+c0. For every a A, g(a) R implies that cna is ∈ ∈ integral over R, therefore cna R and cnA R. The converse is clear. ∈ ⊆ Since a set B K is Rv-fractional if and only if minb B v(b)existsinZ ⊆ ∈ ∪ , Lemma 1.0 shows that A being Rv-fractional is necessary and sufficient for {∞} mina A v(f(a)) to exist in Z for any non-constant f K[x]. To exclude ∈ ∪{∞} ∈ polynomials identically zero on A, for which mina A v(f(a)) = , we need deg f< A, so that the conditions on A in Lemma 1.1 below∈ are necessary.∞ | | 1.1 Lemma. Let n N0.IfAis an Rv-fractional subset of K with A >n,then max min v(f(a)) f∈monic K[x], deg f = n exists. | | {a A | ∈ } ∈ Proof. The case n = 0 is trivial; so let n>0andm Nsuch that A is not m ∈ contained in any union of n cosets of Mv in K. Such an m exists, since n< A m | | and by the Krull Intersection Theorem m Mv = (0). We show that for every ∈N monic f K[x] of degree n there exists an a0 A with v(f(a0))

ring with maximal ideal Mv0 ,ande=[Γv0 :Γv]. A is not contained in any union me me of n cosets of Mv in K0.Pickana0 Athat is not in u + Mv for any root 0 ∈ n 0 u of f in K0;thenv(f(a0)) = v0(f(a0)) = v0(a0 ui)

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Proof. Let In,v = leading coefficients of degree n polynomials in Int(A, Rv) { }∪ 0 . The leading coefficient cn of any n-th degree polynomial in Int(A, Rv)must { } γv,A(n) satisfy v(cn) γv,A(n), so In,v Mv− .Now,forn N0,letgn be monic ≥− ⊆ ∈ of degree n in K[x] with mina A v(gn(a)) = γv,A(n) (such things exist by dint of ∈ γv,A(n) γv,A(n) Lemma 1.1). Then Mv− gn Int(A, Rv), so Mv− In,v. This shows ⊆ ⊆ (i) and also that In,vgn Int(A, Rv) for all n N0. (ii) follows by Lemma 0.1 and γv,A(n)⊆ ∈ the fact that Mv− = cnRv for every cn K with v(cn)= γv,A(n). ∈ − Before deriving a formula for max mina A v(f(a)) f monic K[x], deg f = n , { ∈ | ∈ } when A is a subring of Rv, we check that the other plausible way of normalizing the polynomials would yield the same value. We also see that polynomials mapping A Rv into the maximal possible power of Mv can be chosen to split with their ⊆ roots in any set that Mv-adically approximates A (for instance in A itself, or, if Rv is the localization of a ring R at a prime ideal of finite index, in R). We need a lemma from [7] (but include the proof).

1.3 Lemma. Let f Rv [x], not all of whose coeﬃcients lie in Mv, split over K, ∈ as f(x)=d(x b1) ... (x bm) (x c1) ... (x cl) with v(bi) < 0,v(ci) 0, − · · − · − · · − ≥ and put f+(x)=(x c1) ... (x cl). Then, for all r Rv, v(f(r)) = v(f+(r)). − · · − ∈ m Proof. For r Rv v(r bi )=v(bi)andsov(f(r)) = v(d)+ i=1 v(bi)+v(f+(r)); ∈ m− 1 n n 1 we show v(d)= j=1 v(bi). Consider d− f(x)=x +an 1x − +...+a0.Since − −P f Rv[x] Mv[x], v(d)= min0 k n v(ak). But ak is the elementary symmetric ∈ \ P − ≤ ≤ polynomial of degree n k in the bi and ci, so the minimal valuation is attained m − by v(an m)= i=1 v(bi). −

1.4 Proposition.P Let A Rv and 0 n< A;thenαand γ below are equal: ⊆ ≤ | | α =max min v(f(a)) f Rv[x] Mv[x], deg f = n , {a A | ∈ \ } ∈ γ =max min v(f(a)) f monic K[x], deg f = n . {a A | ∈ } ∈ l If, furthermore, B Rv, such that B intersects every coset of Mv that A intersects, ⊆ for all l N,thenδbelow is equal to α and γ;andsoisβ,ifBis also a ring: ∈ β =max min v(f(a)) f B[x] (Mv B)[x], deg f = n , {a A | ∈ \ ∩ } ∈ n δ =max min v(f(a)) f(x)= (x di),di B. {a A | i=1 − ∈ } ∈ Y Proof. Let B be a fixed subset of Rv that intersects every coset of every power of Mv that A intersects (e.g. B = Rv, when only interested in α and γ). For n = 0 all four expressions are equal to 0; now consider a fixed n>0. Clearly δ γ and, if B is a ring, δ β α.Alsoγ α, because, given f monic in ≤ ≤ ≤ ≤ K[x], there exists a d Rv such that df = g Rv[x] Mv[x]andforalla A ∈ ∈ \ ∈ v(g(a)) = v(d)+v(f(a)) v(f(a)), and so mina A v(g(a)) mina A v(f(a)). ≥ ∈ ≥ ∈ To show α δ,wefixf Rv[x] Mv[x] of degree n and construct a monic ≤ ∈ \ g that splits with roots in B such that v(g(a)) mina A v(f(a)) for all a A. ≥ ∈ ∈ Let v0 be an extension of v to the splitting field of f over K. For all a A, l ∈ v0(f(a)) = v0(f+(a)) with f+(x)= (x ci), where the ci are the roots of f in i=1 − Rv0 , by Lemma 1.3. Put s =mina Av0(f+(a)). We replace each ci by a di B l Q∈ ∈ chosen such that i=1(x di)=h(x)satisfiesv0(h(a)) s for all a A.If k − s≥ ∈ (ci+Mv) A= for all k N,wepickdi out of (ci + Mv ) B; otherwise out of 0 ∩ 6 ∅Q ∈ 0 ∩

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k k (ci + M ) B with k maximal such that (ci + M ) A = . Since the intersection v0 v0 ∩ k ∩ 6 ∅ of a residue class of M in Rv with Rv is either empty or an entire residue class of v0 0 apowerofMv in Rv,andBintersects all of these that A intersects, it is possible to ﬁnd such di in B.Nowforeverya Aeither v0(a di) v0(a ci) for all i and ∈ − ≥ − so v0(h(a)) v0(f+(a)) s,orv0(a di) sfor some i and hence v0(h(a)) s. ≥ ≥ − ≥ n l ≥ To get a polynomial of degree n,setg(x)=(x d0) − h(x), d0 B. − ∈

2. Polynomials mapping into a maximal power of Mv

If R is an infinite subring of a discrete valuation ring Rv, we will construct polynomials gn(x)=(x a1)...(x an)thatmapRinto the maximal possible − − (for a monic polynomial of degree n)powerofMv, by finding sequences (ai)inR n that show a nice distribution among the cosets of Mv R, to serve as roots. This generalizes a procedure of Pólya [16] (also used∩ by Gunji and McQuillan [12], [14], Cahen [4] and others) for the special case where Rv = RQ, Q beingaprime 2 ideal of index q in R such that RQ is a discrete valuation ring: Pick π Q Q ∈ \i and a complete set of residues r0, ..., rq 1 of Q in R and define an = i 0 rci π ,if i − ≥ n= i 0ciq is the q-adic expansion of n. The resulting polynomials map R into ≥ P the highest possible power of Q and can be used to give a regular basis of Int(Rv) (mostP clearly stated in [14]). Gilmer [10] has remarked that the construction even works for Int(D), D a quasi-local ring with principal maximal ideal. The -sequences below are defined for any commutative ring R. All sequences I are indexed by an initial segment of N or N0. Quantifiers over indices of such a sequence are assumed to range over precisely the index-set. 2.0 Definition. If is a set of ideals in a commutative ring R, we define an I -sequence in R to be a sequence (an)ofelementsinRwith the property I I n, m an am mod I [ R : I ] n m. ∀ ∈I ∀ ≡ ⇐⇒ | − We define a homogeneous -sequence to be one with the additional property I I n 1 an I [ R : I ] n. ∀ ∈I ∀ ≥ ∈ ⇐⇒ | (Any infinite [R : I] we regard as dividing 0, but no other integer.) Note that a1,a2,... is a homogeneous -sequence if and only if 0 = a0,a1,a2,... is an -sequence. I I 2.1 Proposition. Let = In n N be a descending chain of ideals in a commutative ring R. ThenI there{ exists| ∈ an} infinite homogeneous -sequence in R. I (k) Proof. Put I0 = R.Fork 0, if [Ik : Ik+1] is finite, let aj 0 j<[Ik:Ik+1] ≥ (k) { | ≤ (k) } be a system of representatives of Ik : Ik+1 with a0 =0,otherwiselet(aj )j 0 ∈N (k) be a sequence in Ik of elements pairwise incongruent mod Ik+1,witha0 =0.If IN with [R : IN ] finite, then every n<[R:IN] has a unique representation ∈IN 1 N 1 (k) n = − j [R : I ]with0 j <[I :I ], and we set a = − a . If the k=0 k k k k k+1 n k=0 jk indices of ideals in get arbitrarily≤ large while remaining finite, this defines our P I P -sequence inductively. Otherwise there exists IN of maximal finite index such I ∈I that either [IN : IN+1] is infinite or Im = IN for m N. Define an for n<[R:IN] (N) ≥ as above. Then, in the first case, set am = aq + ar for m = q [R : IN ]+r with 0 r<[R:IN], and am = ar in the second. ≤

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2.2 Facts. (i) For I of ﬁnite index in R,any[R:I]consecutive terms of an -sequence form a complete∈I set of representatives of R mod I. I n n (ii) If (ai)i=1 is an -sequence in R then (r ai)i=1 is an -sequence for every nI 1 − I r R and (an an i)i=0− is a homogeneous -sequence. ∈ − − I The following lemma will be needed for globalization.

2.3 Lemma. If a1, ..., al is an -sequence for a chain of ideals , J with I I ∈I [R : J] >l,andb1, ..., bl R such that bn an mod J for 1 n l,then(bn)is ∈ ≡ ≤ ≤ also an -sequence, and homogeneous if (an) is. I Proof. Let I and 1 n, m l. First suppose n m mod [R : I]. Then n = m ∈I ≤ ≤ ≡ or [R : I]

2.5 Definition. For n N0 , R an infinite subring of Rv and q N,let ∈ ∈ n n αv,R(n)= and αq(n)= . [R:M j R] qj j 1 v j 1 X≥ ∩ X≥ n Infinite indices are allowed; =0.SinceRis infinite, αv,R(n)isalwaysa ∞ finite number. We will frequently use the fact that αv,R(n) > 0 if and only if n [R : Mv R]. If Q is a prime ideal in a domain D, such that DQ is a discrete ≥ ∩ valuation ring, we write vQ for the corresponding valuation with value group Z.

2.6 Facts. (i) If Q is a prime ideal of ﬁnite index q in R such that RQ is a discrete

valuation ring, then αvQ,R(n)=αq(n) for all n. (ii) If v is a discrete valuation, R an inﬁnite subring of Rv and v0 an extension of v with [Γv :Γv]=eﬁnite, then αv ,R(n)=e αv,R(n) for all n. 0 0 · n n Proof. (i) Since Q is maximal, (QRQ) R = Q for all n. Using the fact that ∩ n n n Q contains a generator of QRQ one sees that [R : Q ]=[RQ :(QRQ) ]=q for k k k e all n. (ii) For k N, Mv0 R =(Mv0 Rv) R=Mvd e R,where x denotes ∈ ∩ ∩ ∩ n∩ d e the smallest integer greater or equal x. Each number j appears e times, [R:Mv R] ∩ as n for k =(j 1)e +1,..., je,inthesumforh α i(n). [R:M k R] v0,R v0 ∩ − h i In the remainder of section two, v is assumed to have value-group Z.

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n+1 n n n 2.7 Lemma. Let (ai)i=1 , (bi)i=1 and (ci)i=1 be v-sequences for R,and(ci)i=1 homogeneous. Then

(a) v(c1 ... cn)=αv,R(n) v(b1 ... bn) αv,R(n)+max1 i nv(bi), ·n · ≤ · · n ≤ ≤ ≤ (b) v (an+1 ai) = αv,R(n) v (r bi) for all r R. i=1 − ≤ i=1 − ∈

Proof. vQ(c1 ... cn)= j 1 i 1 i Qn, v(ci) j and similarly for the bi. · · j ≥ {| ≤ ≤ j ≥ } Since for ﬁnite index Mv R every [R : Mv R] successive terms of a v-sequence P∩ ∩j form a complete residue system of R mod Mv R,wehave j N ∩ ∀ ∈

n n i v(ci) j = j i v(bi) j j +1. { | ≥ } [R:Mv R] ≤ { | ≥ } ≤ [R:Mv R] ∩ ∩ j This implies (a) (and, since the 1 on the right can only occur if [R : Mv R] n, j ∩ 6| v(b1 ... bn) αv,R(n)+max1 i nv(bi) max j [R : Mv R] divides n ). By Fact· 2.2 (ii)· about≤ -sequences,≤ (b)≤ is a special− case{ | of (a). ∩ } I 2.8 Theorem. Let R be an inﬁnite subring of Rv.AnRv-basis of Int(R, Rv) is given by n i=1(x ai) f0 =1 and fn(x)= n − (n 1), (an+1 ai) ≥ Qi=1 − where (an)n∞=1 is a v-sequence for R. Q

Proof. An inﬁnite v-sequence (an)n∞=1 in R exists by Proposition 2.1 applied to n Mv R n N .Thefn, being a K-basis of K[x], are free generators of the { ∩ | ∈ } Rv-module they generate in K[x], call this module F . Since by Lemma 2.7 every fn maps R to Rv, F Int(R, Rv). For the reverse inclusion we show the stronger ⊆ statement that Int(A, Rv) F ,whereA= an n N .Letf Int(A, Rv), f = N ⊆ { | ∈ } ∈ lj fj with lj K. We show inductively that the lj are in Rv. l0 = f(a1) Rv. j=0 ∈ ∈ The induction hypothesis is lj Rv for 0 j

Corollary 1. αv,R(n)=maxmin v(f(r)) fmonic K[x], deg f = n and {r R | ∈ } ∈ αv,R(n) Mv− = leading coefficients of n-th degree polynomials in Int(R, Rv) 0 . { }∪{ } Proof. The second statement can be read off the theorem using Lemma 2.7 (b); the first one then follows by Theorem 1.2. Pólya’s Satz IV [16] is a special case: if P is a prime ideal in a domain R such that RP is a discrete valuation ring and [R : P ]=q, then (by Proposition 1.4 with B = R

and Fact 2.6 i) αq(n)=max minr R vP (f(r)) f R[x] P [x], deg f = n . { ∈ | ∈ \ } n (n) (n) n Corollary 2. Let gn(x)= (x ai ),where(ai ) is a v-sequence for R i=1 − i=1 when n [R : Mv R],andletgn be any monic polynomial in Rv[x] of degree n ≥ ∩ Q

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for 0 n<[R:Mv R]. Further, for n N0,letcn Kwith v(cn)= αv,R(n). ≤ ∩ ∈ ∈ − Then (cngn)n 0 is an Rv-basis of Int(R, Rv). ∈N Proof. For all n N0 , r R, v(g n (r)) αv,R(n) (by Lemma 2.7, in case n ∈ ∈ ≥ ≥ [R : Mv R], and because gn Rv[x]andαv,R(n) = 0 otherwise). By the maxi- ∩ ∈ mality of αv,R(n) (Corollary 1), minr R v(gn(r)) = αv,R(n). Therefore (cngn)n 0 ∈ ∈N is an Rv-basis of Int(R, Rv) by Corollary 1 and Theorem 1.2 (ii).

3. Polynomials mapping a subring into a Krull ring

Notation. Let S be a domain with quotient field K, such that S = v Rv, a ∈V V set of discrete valuations (with value-group Z)onK;andRan infinite subring of T S. We put In = leading coefficients of n-th degree polynomials in Int(R, S) 0 and introduce names{ for recurring additional conditions: }∪{ } (F) q N Q E R [R : Q]=q and Q = Mv R for some v is a finite ∀set.∈ { | ∩ ∈V} n (C) For every prime ideal Q of finite index in R,thesetofMv Rwith n N, ∩ ∈ v ,andMv R=Q, if not empty, forms a descending chain of ideals. ∈V ∩ Note that (C) holds naturally in two cases: when there is only one Mv such that n Mv R = Q, and when every Mv R with Mv R = Q is a power of Q. ∩ ∩ ∩ 3.0 Lemma (Cahen [4]). If R is an infinite subring of a Krull ring S and q N, then S has at most finitely many height 1 prime ideals P with [R : P R]=q∈. ∩ Proof. There exists r R with rq r = 0. For every P with Q = R P of index q in R, rq r Q P ,∈ so the statement− 6 follows by the definition of Krull∩ ring. − ∈ ⊆ 3.1 Lemma. Let v be such that Mv R = Q =(0),andLthe quotient field of ∈V ∩ 6 R.IfRQis a valuation ring, then it is a discrete valuation ring and RQ = Rv L. n ∩ If Q is also a maximal ideal, then, for every n N, Mv R is a power of Q. ∈ ∩ Proof. For any valuation ring V with quotient field L and maximal ideal M we 1 have L V = r L∗ r− M .PutRv L=Rwand Mv L = Mw;then \ { ∈ | ∈ } ∩ ∩ Rw and RQ are valuation rings with quotient field L and maximal ideals Mw and QRQ, respectively. R Rw and Mw R = Mv R = Q imply RQ Rw and ⊆ ∩ ∩ 1 ⊆ also QRQ Mw. By the latter inclusion L RQ = r L∗ r− QRQ ⊆ 1 \ { ∈ | ∈ }⊆ r L∗ r− Mw =L Rw. This shows RQ = Rw = Rv L,soRQ is a { ∈ | ∈ } \ n ∩ discrete valuation ring and every Mv RQ is a power of QRQ.IfQis maximal, k k ∩n then (QRQ) R = Q for all k,soM Ris a power of Q. ∩ v ∩ 3.2 Lemma. (C) implies: For every finite set of prime ideals of finite index in M m R and every m N, there exists a sequence (ai)i=0 in R that is a homogeneous ∈ v-sequence for all v in with Mv R , simultaneously. V ∩ ∈M n Proof. For every Q , Q = Mv R v ,n N,Mv R = Q (if not ∈M I { ∩ | ∈V ∈ ∩ } empty) is a descending chain by (C), so there exists a homogeneous Q-sequence (Q) I (ai )i∞=0 in R by Proposition 2.1. For each Q with Q = let IQ be an element of n I 6 ∅ n Q with [R : IQ] >m.IQ =Mv Rfor some v and n, and therefore it contains Q . SinceI different Q are co-prime, there∩ exists, by the Chinese Remainder Theorem, a m (Q) m sequence (ai) in R that is congruent to (ai ) modulo IQ for all Q .By i=0 i=0 ∈M Lemma 2.3, this a homogeneous Q-sequence for all Q , i.e., a homogeneous I ∈M v-sequence for all v with Mv R . ∩ ∈M

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From Lemma 3.0, Lemma 3.1 and the fact that the powers of an ideal Q form a descending sequence, we conclude that the hypothesis of Theorem 3.4 below is satisﬁed in at least one natural setting:

1 3.3 Fact. If S is a Krull ring, = vP P Spec (S) ,andRan inﬁnite subring V { | ∈ } 1 such that RQ is a valuation ring for every ﬁnite index Q = P R, P Spec (S), then (C) and (F) both hold. ∩ ∈ In the following theorem, the case where S is a Dedekind ring and R = S is due to Cahen [4] (also pertinent: [5]).

3.4 Theorem. Let R be an inﬁnite subring of S = v Rv.If(C) and (F) hold, then ∈V T αv,R(n) In = Mv− (n N0) ∈ v \∈V and there exists a regular sequence of monic polynomials (gn) in R[x] such that

Int(R, S)= Ingn, n 0 X≥ n (n) (n) n namely, gn(x)= (x ai ),where(ai ) is a simultaneous v-sequence for i=1 − i=1 all v with [R : Mv R] n. ∈V Q ∩ ≤ αv,R(n) Proof. Int(R, v Rv)= v Int(R, Rv), therefore In v Mv− (by ∈V ∈V ⊆ ∈V αv,R(n) Theorem 2.8, Corollary 1). For the reverse inclusion, let c v Mv− .Set T T ∈ T ∈V n= v αv,R(n) > 0 = v [R:Mv R] n ;then Mv R v n V { ∈V| (n)}n { ∈V| ∩ ≤ } T { ∩ | ∈V } is ﬁnite by (F). Let (ai )i=1 in R be a homogeneous v-sequence for all v n n (n) ∈V simultaneously (which exists by Lemma 3.2) and gn(x)= (x ai ). Then i=1 − minr R v(g(r)) αv,R(n) for all v (by Lemma 2.7 when v n, and because ∈ ≥ ∈V Q ∈V αv,R(n)=0andgn R[x] otherwise), which means cg(x) Int(R, v Rv)and ∈ ∈ ∈V hence c In. This completes the proof of the ﬁrst statement and also shows, for ∈ T all n 0, that Ingn Int(R, v Rv), so the second follows by Lemma 0.1. ≥ ⊆ ∈V From now on, S is a Krull ring.T By convention, the empty intersection or product of ideals of S equals S. We denote the set of height 1 prime ideals of S by Spec1(S) (j) j or .IfP ,wewriteαP,R for αvP ,R and, if j N0, P for (PSP) S.With P ∈P ∈ ∩ this notation we have, for n N0 and P : ∈ ∈P n α (n)= . P,R [R:P(j) R] j 1 X≥ ∩

3.5 Lemma. Let S be a Krull ring and = vP P .If(C) holds, then V { | ∈P}(αP,R(n)) Int(R, S) has a regular basis if and only if P ,[R:P R] n P is principal for all n. ∈P ∩ ≤ T Proof. αP,R(n) =0ifandonlyif[R:P R] n. Since (F) holds by Lemma 3.0, 6 ∩ ≤ this only happens for ﬁnitely many P for each n.IfaP P is a set of { | ∈P}aP integers, only ﬁnitely many of them non-zero, then P (PSP)− is principal if aP ∈P and only if P (PSP) is, namely if there exists c K with vP (c)=aP for all ∈P Ta ∈ P P .Ifalla are non-negative then (PS ) P = P(aP ). Applied to P P P aP >0 ∈ T ∈P (PS ) αP,R(n),whichisI by Theorem 3.4, with Lemma 0.1 (iii) in mind, P P − n T T this∈P proves the claim. T

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3.6 Theorem. Let R be an inﬁnite subring of a Krull ring S, =Spec1(S), P ∗= P [R:P R]ﬁnite and = R P P ∗ .IfRQis a valuation P { ∈P| ∩ } Q { ∩ | ∈P } ring for all Q ,thenRQ is a discrete valuation ring for all Q and ∈Q ∈Q

(eP ) Int(R, S) has a regular basis q N P is a principal ideal of S, ⇐⇒ ∀ ∈ P [R:R\∈PP ]=q ∩

where eP is the ramiﬁcation index of PSP over QRQ,forP ∗,Q=P R. ∈P ∩ Proof. Let q = P [R:P R]=q ,P q,Q=P R,Lthe quotient P { ∈P| ∩ } ∈P ∩ ﬁeld of R; then by Lemma 3.1 RQ = SP L and RQ is a discrete valuation ring. v =(1/e )v is equivalent to v and is an∩ extension of v to K with [Γ :Γ ]= P0 P P P Q vP0 vQ e . By the Facts 2.6 (ii) and (i), α (n)=α (n)=e α (n)=e α (n). P P,R vP0 ,R P Q,R P q If we call the left and right sides of the claimed equivalence (l) and (r), re- (αP,R(n)) spectively, then (l) is equivalent to (l0)‘ n P P is principal’ by ∀ [R:P ∈PR] n ∩ ≤ Lemma 3.5 (whose condition (C) holds by FactT 3.3). We know that P (αP,R(n)) = P (eP αq (n)).

P q n P q [R:P\∈PR] n \≤ \∈P ∩ ≤ The latter is clearly principal provided all P (eP ) are; thus (r) (l ). P q 0 ∈P ⇒ For (l) (r), suppose P (eP αq (n)) = s S for all n.Weseethat q n P q n ⇒ ≤ ∈P T sS= P (eP ) P (eP αl(q)), because α (q) = 1. This allows an q P q l

Int(R, S) has a regular basis q N div(QS) is principal, ⇐⇒ ∀ ∈ Q Spec1(R) ∈[R:\Q]=q where div(QS) means the smallest divisorial ideal containing QS. Proof. If R S is an extension of Krull rings with the stated property and Q 1 ⊆ (eP ) is in Spec (R), then div(QS)= P Spec1(S) P ,whereeP =e(PQ)isthe ∈P R=Q | ∩ ramiﬁcation index of PSP over QRTQ [1, p. 183]. In particular, if R S is an extension of Dedekind rings, then ⊆ Int(R, S) has a regular basis q N QS is principal. ⇐⇒ ∀ ∈ Q Spec(R) [∈R:YQ]=q A diﬀerent specialization gives Ostrowski’s criterion [15]. If S is a Krull ring,

Int(S) has a regular basis q N P is principal. ⇐⇒ ∀ ∈ P Spec1(S) ∈[S:YP ]=q When a regular basis exists, we can give a fairly explicit description of one. (For Int(S), S a Dedekind ring, there also is a diﬀerent construction by Gerboud [9].)

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(eP ) Corollary 2. In the situation of Theorem 3.6, if P P = cqS (q N) [R:P ∈PR]=q ∈ ∩ then a regular basis of Int(R, S) is given by f0 =1T, n αq(n) (n) fn(x)= c−q (x ai )(nN) − ∈ q n i=1 Y≤ Y (n) n where (ai ) R is a vP -sequence for all P with [R : P R] n. i=1 ⊆ ∈P ∩ ≤ αq (n) Proof. vP (c−q )= ePαq(n)= αP,R(n)fortheP with [R : P R]=q, − − αq(n) ∈P ∩ and zero for all other P ,sovP( q nc−q )= αP,R(n) for all P (since ∈P ≤ − ∈P αP,R(n)=0ifn<[R:P R]). Therefore the fn are an SP -basis of Int(R, SP )for all P simultaneously,∩ by TheoremQ 2.8, Corollary 2. ∈P References

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Institut fur¨ Mathematik C, Technische Universitat¨ Graz, Kopernikusgasse 24, A- 8010 Graz, Austria E-mail address: [email protected]

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